Show that the condition number of an invertible matrix must be at least 1. What matrices have
condition number equal to 1.
If someone could help me with this and give an explanation that would be very helpful. I do not know where to start
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Hint: note that $\|AB\| \leq \|A\|\cdot \|B\|$ (that is, $\|\cdot\|$ is "sub-multiplicative").
So, $\|A\|\cdot \|A^{-1}\| \geq \cdots ?$
Ben Grossmann
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Mind your formatting, but yes. Now: what matrices have condition number equal to $1$? – Ben Grossmann Mar 31 '14 at 02:36
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Okay, so what kind of matrices $A$ satisfy $|A| \cdot |A^{-1}| = 1$? We can be a bit more specific. – Ben Grossmann Mar 31 '14 at 03:00
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When the reciprocal of the norm of the inverse of A is equal to the norm of A? – Vogtster Mar 31 '14 at 03:01
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Again, you're not saying anything different. Come to think of it, it's not clear what kind characterization they're looking for. However, it is true that $|A| \cdot |A^{-1}| = 1$ if and only if all the singular values of $A$ are identical – Ben Grossmann Mar 31 '14 at 03:03
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So if all the singular values are the same, which is the same to say that all the eignvalues are the same magnitude? – Vogtster Mar 31 '14 at 03:04
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Not necessarily. We can say, however, that $A$ must be a scalar multiple of a unitary (or real orthogonal) matrix. That is, a non-zero $A$ has this property if and only if its columns (or rows if you prefer) are of the same length mutually orthogonal. – Ben Grossmann Mar 31 '14 at 03:07
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Just curious, youve been very helpful by the way. The original hint you gave, is this like a generalized cauchy shawtz? – Vogtster Mar 31 '14 at 04:08
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@user136693 You could certainly look at it that way! The way I see it, though, it is one of the defining properties of a matrix norm – Ben Grossmann Mar 31 '14 at 11:51
A^{-1}results in $A^{-1}$. – Emily Mar 31 '14 at 03:01